Question

A common laboratory preparation for small for quantities of `O_2` is to decompose `KClO_3` by heating: `2KClO_3 (s) -> 2KCl (s) + 3O_2 (g)`. If the decomposition of 2.00 g of `KCIO_3` gives 0.720 g of `O_2`, what is the percent yield for the reaction?

Answer 1

`91.9%`

Explanation:

Start by examining the balanced chemical equation that describes this decomposition reaction

`color(darkgreen)(2)"KClO"_ (3(s)) stackrel(color(white)(acolor(red)(Delta)aaa))(->) 2"KCl"_ ((s)) + color(blue)(3)"O"_ (2(g))`

Notice that it takes `color(darkgreen)(2)` moles of potassium chlorate to produce `color(blue)(3)` moles of oxygen gas. This represents the reaction's theoretical yield, i.e. the amount of oxygen gas that you can expect to see for a reaction that has a `100%` yield.

The reaction provides you with grams of potassium chlorate, so use the molar masses of potassium chlorate and of oxygen gas to convert the mole ratio to a gram ratio.

`(color(darkgreen)(2)color(red)(cancel(color(black)("moles KClO"_3))) * "122.55 g"/(1color(red)(cancel(color(black)("mole KClO"_3)))))/(color(blue)(3)color(red)(cancel(color(black)("moles O"_2))) * "32.0 g"/(1color(red)(cancel(color(black)("mole O"_2))))) = "2.553 g KClO"_3/"1 g O"_2`

This means that for every `"2.553 g"` of potassium chlorate that undergo combustion, the reaction can theoretically produce `"1 g"` of oxygen gas.

Now, the problem tells you that you decompose `"2.00 g"` of potassium chlorate. The reaction should theoretically produce

`2.00 color(red)(cancel(color(black)("g KClO"_3))) * "1 g O"_2/(2.553color(red)(cancel(color(black)("g KClO"_3)))) = "0.7834 g O"_2`

This represents the reaction's theoretical yield. You know that the reaction actually produced `"0.720 g"` of oxygen gas, which means that its percent yield was

`(0.720 color(red)(cancel(color(black)("g O"_2))))/(0.7834color(red)(cancel(color(black)("g O"_2))))xx 100 = color(green)(|bar(ul(color(white)(a/a)color(black)(91.9%)color(white)(a/a)|)))`

The answer is rounded to three sig figs.